The expression 6x^2 + bx - 35, where b is a constant, can be rewritten as \((\mathrm{h}\mathrm{x} + \mathrm{k})(\mathrm{x} +...
GMAT Advanced Math : (Adv_Math) Questions
The expression \(6\mathrm{x}^2 + \mathrm{b}\mathrm{x} - 35\), where \(\mathrm{b}\) is a constant, can be rewritten as \((\mathrm{h}\mathrm{x} + \mathrm{k})(\mathrm{x} + \mathrm{j})\), where \(\mathrm{h}\), \(\mathrm{k}\), and \(\mathrm{j}\) are integer constants. Which of the following must be an integer?
1. INFER the key strategy
- We need to find what must always be an integer when \(6\mathrm{x}^2 + \mathrm{b}\mathrm{x} - 35 = (\mathrm{h}\mathrm{x} + \mathrm{k})(\mathrm{x} + \mathrm{j})\)
- Key insight: Expand the right side and compare coefficients to create relationships
2. SIMPLIFY by expanding the factored form
- Expand \((\mathrm{h}\mathrm{x} + \mathrm{k})(\mathrm{x} + \mathrm{j})\):
\((\mathrm{h}\mathrm{x} + \mathrm{k})(\mathrm{x} + \mathrm{j}) = \mathrm{h}\mathrm{x}^2 + \mathrm{h}\mathrm{j}\mathrm{x} + \mathrm{k}\mathrm{x} + \mathrm{k}\mathrm{j} = \mathrm{h}\mathrm{x}^2 + (\mathrm{h}\mathrm{j} + \mathrm{k})\mathrm{x} + \mathrm{k}\mathrm{j}\)
3. INFER relationships by comparing coefficients
- Comparing \(\mathrm{h}\mathrm{x}^2 + (\mathrm{h}\mathrm{j} + \mathrm{k})\mathrm{x} + \mathrm{k}\mathrm{j}\) with \(6\mathrm{x}^2 + \mathrm{b}\mathrm{x} - 35\):
- Coefficient of \(\mathrm{x}^2\): \(\mathrm{h} = 6\)
- Coefficient of \(\mathrm{x}\): \(\mathrm{h}\mathrm{j} + \mathrm{k} = \mathrm{b} \rightarrow 6\mathrm{j} + \mathrm{k} = \mathrm{b}\)
- Constant term: \(\mathrm{k}\mathrm{j} = -35\)
4. INFER the constraint on k
- Since \(\mathrm{k}\mathrm{j} = -35\) and both k and j must be integers
- k must be a divisor of -35
- Possible values for k: \(\pm 1, \pm 5, \pm 7, \pm 35\)
5. INFER which expression must be an integer
- Check each option:
- (A) \(\mathrm{b}/\mathrm{h} = (6\mathrm{j} + \mathrm{k})/6 = \mathrm{j} + \mathrm{k}/6\) ← \(\mathrm{k}/6\) not always integer
- (B) \(\mathrm{b}/\mathrm{k} = (6\mathrm{j} + \mathrm{k})/\mathrm{k} = 6\mathrm{j}/\mathrm{k} + 1\) ← \(6\mathrm{j}/\mathrm{k}\) not always integer
- (C) \(35/\mathrm{h} = 35/6\) ← not an integer
- (D) \(35/\mathrm{k}\) ← Since k divides -35, it also divides 35, so this IS always an integer
Answer: D
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students may try to work backwards from the answer choices without establishing the fundamental relationships first.
They might substitute specific values for the variables or try algebraic manipulation on the answer choices without recognizing that they need to expand and compare coefficients first. This leads to confusion about what constraints actually exist in the problem.
This leads to confusion and guessing among the answer choices.
Second Most Common Error:
Inadequate SIMPLIFY execution: Students correctly identify that they need to expand \((\mathrm{h}\mathrm{x} + \mathrm{k})(\mathrm{x} + \mathrm{j})\) but make errors in the algebraic expansion or fail to properly organize terms for coefficient comparison.
Without the correct expansion \(\mathrm{h}\mathrm{x}^2 + (\mathrm{h}\mathrm{j} + \mathrm{k})\mathrm{x} + \mathrm{k}\mathrm{j}\), they can't establish that \(\mathrm{h} = 6\), \(\mathrm{b} = 6\mathrm{j} + \mathrm{k}\), and \(\mathrm{k}\mathrm{j} = -35\). This prevents them from recognizing the divisibility constraint.
This may lead them to select Choice A or B by attempting to analyze expressions involving b without the proper relationships.
The Bottom Line:
This problem requires students to bridge polynomial algebra with number theory concepts - they must expand correctly AND recognize divisibility implications. The key insight is that the constraint \(\mathrm{k}\mathrm{j} = -35\) forces k to be a divisor of 35, making \(35/\mathrm{k}\) the only expression guaranteed to be an integer.